- #1

- 371

- 0

[tex]\frac{{\partial}^{2}f}{{\partial x}^{2}}+\frac{{\partial}^{2}f}{{\partial y}^{2}}=0[/tex]

Thanks in advance.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter dimension10
- Start date

- #1

- 371

- 0

[tex]\frac{{\partial}^{2}f}{{\partial x}^{2}}+\frac{{\partial}^{2}f}{{\partial y}^{2}}=0[/tex]

Thanks in advance.

- #2

- 473

- 146

You really have to give the boundary conditions when dealing with Laplace equation, and in general for PDEs, boundary and initial conditions.

- #3

- 1,796

- 53

The general solution is:

[tex]f(x,y)=g(z)+h(\overline{z}),\quad z=x+iy[/tex]

where g and h are arbitrary [itex]C^2[/itex] functions of a single variable z=x+iy. We can show this by factoring the equation:

[tex]\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right) f=\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)=\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)u=0[/tex]

where:

[tex]u=\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}[/tex]

So first solve:

[tex]u_t-iu_x=0[/tex]

which is an easy first-order equation which turns out to be u=g(x+it) then substitute that into:

[tex]f_x+if_y=g(x+it)[/tex]

another easy one to obtain so that the general solution is:

[tex]f(x,y)=g(x+it)+h(x-it)[/tex]

Note that means the general solution is not analytic since it contains [itex]\overline{z}[/itex] but analyticity is not a requirement for the solution but only differentiability.

[tex]f(x,y)=g(z)+h(\overline{z}),\quad z=x+iy[/tex]

where g and h are arbitrary [itex]C^2[/itex] functions of a single variable z=x+iy. We can show this by factoring the equation:

[tex]\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right) f=\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)=\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)u=0[/tex]

where:

[tex]u=\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}[/tex]

So first solve:

[tex]u_t-iu_x=0[/tex]

which is an easy first-order equation which turns out to be u=g(x+it) then substitute that into:

[tex]f_x+if_y=g(x+it)[/tex]

another easy one to obtain so that the general solution is:

[tex]f(x,y)=g(x+it)+h(x-it)[/tex]

Note that means the general solution is not analytic since it contains [itex]\overline{z}[/itex] but analyticity is not a requirement for the solution but only differentiability.

Last edited:

- #4

- 371

- 0

The general solution is:

[tex]f(x,y)=g(z)+h(\overline{z}),\quad z=x+iy[/tex]

where g and h are arbitrary [itex]C^2[/itex] functions of a single variable z=x+iy. We can show this by factoring the equation:

[tex]\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right) f=\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)=\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)u=0[/tex]

where:

[tex]u=\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}[/tex]

So first solve:

[tex]u_t-iu_x=0[/tex]

which is an easy first-order equation which turns out to be u=g(x+it) then substitute that into:

[tex]f_x+if_y=g(x+it)[/tex]

another easy one to obtain so that the general solution is:

[tex]f(x,y)=g(x+it)+h(x-it)[/tex]

Note that means the general solution is not analytic since it contains [itex]\overline{z}[/itex] but analyticity is not a requirement for the solution but only differentiability.

Thanks.

Share:

A
Laplace or Fourier Transform to solve a system of partial differential equations in thermoelasticity

- Last Post

- Replies
- 2

- Views
- 2K

- Replies
- 0

- Views
- 2K